Cool Math Tricks

After you are through with these tricks you will be hailed as a math genius.

These cool math tricks would help parents help their kids getting up to speed in learning math. Also, a nice and funny way to impart education in your children and pamper their curiosity for numbers.

These cool math tricks, derivation of various vedic math sutras, make calculations possible at your fingertips. Please feel free to leave your comments in case you have any suggestions for additions to this list of cool math tricks! These cool math tricks are particularly important to those appearing and preparing for competitive exams like GMAT, GRE, etc.

Trick to multiply with 11
Multiply any number with 11 easy and fast 

This cool math trick is about multiplication. Not just any number with any other. This math trick has specific application and its about multiplication of any number with 11. With this trick, multiplication of any number with 11 can be calculated in two seconds. Let us see how this trick works with an example of two digit number to be multiplied with 11:

Example:

24 x 11 (Since this math trick is about multiplication with 11, the number we want to multiply 11 with, will be referred to as "Number")

2 _ 4 (The number is split in half from the middle with space in between)

2 6 4 (In the space left in between, total of the two digits on either sides is to be included)

24 x 11 = 264


Cool Math Trick to multiply with 9
use your fingers to multiply any number with 9

This cool math trick is a simple exercise and kids will love this cool math trick. Let us assume, you want to multiply 2 by 9 and don't want to use any of the mathematics concepts. This math trick would let you do it, like this:

1. Raise your both hands up and place them on a platform (anything in front of you, even the desk on which this computer is placed would do). Also, please keep your fingers spread, because that only will make the difference. 
2. As you want to multiply with 2, you just need to fold down the second finger from the left. And just watch your resulting fingers. They would give you the answer.
3. There would be only one finger to the left of folded finger, which signifies that left hand side digit of the answer would be 1. 
4. There are eight finger to the right of folded one, which means the right hand side digit of the answer is 8. (3 of left hand and 5 of right hand) 

Please try this on for any number and this math trick would give you exact right answer for multiplication with 9.

Isn't it cool to know such math tricks! Kids, enjoy playing with these cool math tricks!

Magic with mathematics
let us explore the magic of 1089

This cool math trick would turn you into a magician (Shh! only till your friends also know this :-))

Here is the math trick you need to play with your friends. Why dont you play along as you read. Take a pencil and paper and let us begin with this cool math trick.

I can guess the final answer of all the following steps for any number your guess. And the answer would be 1089!

Isn't it a cool math trick? Enjoy fooling your friends and also think about the reason why any three digit number if passed through following fives steps would give the same result?

The hint to understand logic behind this cool math trick is - "when you reverse a three digit number, tenths place digit would be unmoved and the hundredth and unit would change places"

1. Please write down any three digits number with decreasing digits
   (i.e. 543 or 875, you can also use the number which have zeroes in their tenth or unit place)
2. Reverse the number you wrote in #1
   (If you presumed number with zeros either in tenth or unit or both places, the number you would get in this step would not necessarily be a three digit number)
3. Subtract the number obtained in #2 from the number you wrote in #1
   (Please note, because number in #1 was with digits decreasing, the number got in #2 would be lower)
4. Reverse the number obtained in #3
   (The digits of the subtraction received in #3 to be written in reverse order. Please note, in both the reversals, tenths place digit would be unmoved and the hundredth and unit would change places)
5. Add the numbers found in #3 and #4

Solve complex quadratic equations easily

Solve equations easily and fast

The vedic maths sutra to be used: "Adjust After Transposing (Paravartya Yojayet)"

When students are introduced to the concept of solving equations, the first method taught for solving basic equations is as follows:

Let us say we want to solve the equation 3x - 9 = 0

3x - 9 = 0

3x = 9 (transposing 9 to the other side of equation)

x = 9 / 3 (adjusting 3 with 9 on the right hand side of the equation)

x = 6 (Solution of equation)

This method is little lengthy and can not be used quickly for solving complex equations. Vedic mathematics has a sutra for solving equations and is based on the same concept of transposing and then applying. (Paravartya = Transpose; Yojayet = Adjust)

So let us understand this vedic math trick to solve equations fast and easily.

There is a precondition for using this formula though, and it is that there has to be a linear relationship with ratio of coefficient of one of the unknown in the equation with the constant term. For all the linear equations satisfying this condition, can be solved by simply putting the other unknown (other than the one with ratio of coefficients in proportion of the constant) as zero.

Let us just analyse our normal method of solving equations and see how it actually gives us an answer.

Let us say, we want to solve an equation

ax + b = px + q

ax - px = q - b (transposing both the sides)

x (a - p) = q - b

x = (q - b) / (a - p) (adjusted both the sides)

Thus, every equation that we want to solve using this lengthy method, can be solved easily just by using the final solution that x = (q - b) / (a - p)

Example 1:

5x + 8 = 3x + 16

Without getting into any lengthy calculation, let us solve this equation by using the vedic math trick, the answer would be:

x = (16 - 8) / (5 - 3)

x = 4

Example 2:

4x - 6 = 3x + 16

x = (16 + 6) / (4 - 3)

x = 22

Isn't it simple solving equations using this method instead of getting into length calculations.

Now let us consider solving tricky equations with the same method.

1. Solving equations with undefined solutions

Example 3:

4x + 6 = 4x + 16

x = (16 - 6) / (4 - 4)

x = Undefined (This equation involved division by 0, hence the solution to this equation is undefined as it should be)

2. Solving equations expressed as ratio to one another

While solving equation expressed as ratio to one another, the solution as per vedic math trick would change a little bit as follows:

(ax + b) / (px + q) = y / z

x = (yq - zb) / (za - yp)

Example 4:

(3x + 5) / (6x + 4) = 7 / 8

x = (7*4 - 8*5) / (8*3 - 7*6)

x = (28 - 40) / (24 - 42)

x = (-12) / (-18)

x = 2 /3 (This solution can be checked by trying the traditional method)

Similarly, if (6x + 4) / (3x + 2) = 2, then the same vedic math trick can be used to solve this formula just by replacing 2 with (2/1) on the right hand side of the equation.

When faced with equations such as (2x + 3) / x = 13 / 5, the same vedic math trick can be used and we just need to rewrite the given equation in the format we have the solution, refer the below example.

Example 5:

(2x + 3) / x = 13 / 5

(2x + 3) / (1x + 0) = 13 / 5 (Rewritten the equation to match with the vedic math trick)

x = (13*0 - 5*3) / (5*2 - 13*1)

x = (0 - 15) / (10 - 13)

x = (-15) / (-3)

x = 5 (This solution can be checked by trying the traditional method)

Thus, any equation with single power of the unknown can be easily solved using this vedic math tricks. We just have to make sure to rewrite the equation in the structure provided by the vedic math trick adjust after transposing.

Solving equations is so easy ans fast with the help of vedic math tricks!

Trick to identify exact squares

While vedic math tricks would make squaring numbers very easy, these few tricks help identify perfect squares merely by observing the number.:


Structure of the square root:
First trick is to understand what would be the structure of the square root just by observation.

• The number of digits before decimal point would reduce to half when we take the square root of the number.(i.e. the square root of 1446.78 would have 2 digits before decimal number). This math trick would be applicable directly when the numbers of digits before decimal point are even.
• In case of odd number of digits before decimal point, the square root would still have half of the numbers but that half needs to be rounded up to the next integer. (i.e. square root of 144.678 would have two digits before the decimal point.)
• The number of digits after the decimal point would reduce to half when we take the square root of the number.(i.e. the square root of 1446.78 would have 1 digits after decimal number). This math trick would be applicable directly when the numbers of digits before decimal point are even.
• For numbers with odd digits after decimal point, the square root would be an irrational number, because any square of a whole rational number would have even digits only after the decimal point.

Nature of the square root:Next trick is to know the type of answer to expect while trying to derive square root of a number, the rule is:

• For the numbers not exact squares of another whole number, their square root is always irrational. (i.e. the square root would have endless digits after decimal point and would not have any repeating patters in those digits.)
• Also, a number is odd number of zero's at the end, would never have a rational square root
• For the rest of the numbers the square root would be rational.

The last digit of the answer:This is a trick to identity an exact square and also to estimate the last digit of the square root.

• Number with last digits as 2, 3, 7 or 8 are not exact squares and inturn would result in an irrational number when their square root is calculated
• Also, there is a direct relation between last digit of the number and last digit of its square root, as follows:
• number 1, square root 1 or 9
• number 4, square root 2 or 8
• number 5, square root 5
• number 6, square root 4 or 6
• number 9, square root 3 or 7
• When the second last digit (from right) is even and the last digit is 6, the number is not a perfect square (i.e. 346 is not a perfect square)
• When the second last digit (from right) is odd, the last digit has to be 6 for the number to be a perfect square. (i.e. numbers ending with 34, 59, 11 are not perfect squares)
• When a number is even, and its last two digits taken together are not divisible by 4, that number is not a perfect square. (i.e. numbers ending with 42, 86, etc. are not perfect squares)

Odd one out:This trick will tell us whether the square root would be even or odd.

1. Square root is odd when a perfect square is odd
2. Square root is even when a perfect square is even

These tricks would be particularly helpful when attempting multiple choice questions or math puzzles relating to squares or square roots.

Solve simultaneous linear equations easily

The vedic maths sutra to be used: "If one is in ratio, the other one is zero"

While trying to solve simultaneous equations, especially with bigger numbers, you can use this vedic math sutra to simplify the process.

There is a precondition for using this formula though, and it is that there has to be a linear relationship with ratio of coefficient of one of the unknown in the equation with the constant term. For all the linear equations satisfying this condition, can be solved by simply putting the other unknown (other than the one with ratio of coefficients in proportion of the constant) as zero.

Example:

5x + 8y = 24

23x + 16y = 48

In these equations, we have two unknown (x & y), so let us test if any of them satisfy the required condition:

Testing for x

Ratio of coefficients of x = 23 / 5 = 4.6

Ratio of constant number = 48 / 24 = 2

Not satisfied for x.

Testing for y

Ratio of coefficients of y = 16 / 8 = 2

Ratio of constant number = 48 / 24 = 2

Satisfied for y.

Therefore, the solutions to this equation is x = 0 and therefore y = 3.

however, if both the unknown satisfy the condition, then the equations can not be solved.

The Logic:

The equations consist of two sides LHS and RHS.

Now, RHS is a constant number the ratio of increase is RHS is obvious and can be understood just by observation, which in this case is 2.

This reflects that the LHS of second equation has to deliver a value exactly double as compared to what LHS to first equation delivered.

Now, LHS comprises of two terms with unknown x & y and having different coefficients. For LHS to second equation to become exactly double as compared to the first, both the coefficients should double. But in this case, y is exactly doubling and x is increasing randomly, which clearly means that whatever be the coefficient of x, it has to be zero so as not to disturb the proportion.

Now enjoy solving simultaneous liner equations just by looking at them with the use of this vedic math sutra.

Solve quadratic equations easily with vedic maths

The vedic math sutra used for simplifying equations is : "It is zero if the Samuccaya is Same"

This formula aims to simplify the solution of quadratic equations by setting some simple ground rules. But, before we understand the application of this vedic math sutra with quadratic equation, let us start with basic equations and understand the applicability of this vedic mathss sutra.

This vedic math sutra (It is zero if the Samuccaya is same) has applicability in four different situations:

1. When a common term is there on all the terms of left hand side as well as right hand side of an equation, then that equation can be solved by assuming that common term to be zero.

Example:

11x + 5x = 6x + 7x

Since "x" occurs as a common factor in all the terms (of both LHS & RHS), therefore, x = 0 is a solution.

2. When multiplication of all the numbers appearing on each side of the equation is equal, the solution to that equation can be derived by assuming the unknown to be zero. 

Example:

(x + 6) (x + 5) = (x + 3) (x + 10)

6 * 5 = 3 * 10 (multiplication of each side's numbers is equal = 30)

Solution is x = 0

3. The sum of the fractions with same numerical numerator can be derived by adding the denominators and equating it to zero.

Example:

3/ (4x − 1) + 3/ (6x − 1) = 0

In this case the numerator of both the fractions is the same number 3, therefore the addition can be re-written as:

10x-2 = 0 (LHS simplified by adding the denominators)

x = 1/5

4.  For a quadratic equation expressed as N1/D1 = N2/D2 and if N1 + N2 = D1 + D2, then this sum (i.e. either N1+N2 or D1+D2) is zero. This would help solve the quadratic equation easily.

Example:

(3x + 8) / (3x + 4) = (3x + 4) / (3x + 8)

In this case, the condition can be assessed as:

N1 + N2 = 3x + 8 + 3x + 4 = 6x + 12

D1 + D2 = 3x + 4 + 3x + 8 = 6x + 12

N1 + N2 = D1 + D2

Therefore,

N1 + N2 = D1 + D2 = 0

6x + 12 = 0

x = -2 (Solution to quadratic equation)

Now with the knowledge of this vedic math sutra solving quadratic equation faster is very easy.

Subtracting from a large power of ten

The vedic math sutra - "All from nine and the last from ten"

Subtraction is not a very complex mathematics calculation but it would get difficult when you are dealing with large numbers and even more difficult when it results in carrying over from the left hand site number.

Let us understand where to effectively use this vedic maths sutra:

100000 - 87459

The normal method of subtraction would require us to write down these numbers and do the calculations as follows:

 100000

   87549

   9 9 9 910

 100000

   87549 (carrying over from the left hand side number)

   12451

What if there is a vedic math sutra which shows how to do this calculation without even having to write to down numbers and also without the hustle of having to carry over from the left. Let us lean about the vedic math sutra:

The sutra: All from nine and the last from ten

Method: As the name suggests, you just need to subtract all the numbers from nine and the last one from 10

So in this case, last number is 9 which is to be subtracted from 10 = 1
and the rest of the numbers 8754 to be subtracted from 9 = 1245

So the answer is 12451

Thus, though subtraction is easy, this vedic math sutra makes it even easier!

What is vedic maths

While wonders that vedic mathematis can do in simplifying maths have been discussed at length in Vedic Mathematics, this post is intended for background about vedic maths. Knowing about vedic maths (history, origination) would make it more interesting and appealing.

Vedic Mathematics was rediscovered from the Vedas (Indian Spiritual and knowledge books) between 1911 and 1918 by Sri Bharati Krsna Tirthaji (1884-1960). According to his research all of mathematics is based on sixteen Sutras (formulas). Each of these vedic maths sutras (formulae) describe easier method of how to carry out mathematical calculations mentally and offer best value to students in cracking through exams with high scores.

The most interesting thing about vedic maths is that this method is that it is fully integrated and offers both way proofs of all concepts. meaning multiplication methods can be reversed to use for one-line divisions and the simple squaring method can be reversed to get square roots.

So complex mathematics problems often be solved immediately is you know about vedic maths. These striking and beautiful methods are just a part of a complete system of mathematics which is far more systematic than the modern 'system'. Vedic Mathematics manifests the coherent and unified structure of mathematics and the methods are complementary, direct and easy.

The simplicity of Vedic Mathematics implies easier and mental calculations. There are many advantages in using a flexible, mental system. Students can invent their own methods, they are not limited to the one 'correct' method. This leads to more creative, interested and intelligent learning and would make mathematical concepts easier to grasp for students.

With increasing focus on competitive exams with requirement of faster calculation speeds and, focus on Vedic maths system is growing in education. Lot of new research is being carried out to augment established vedic maths sutras and also enhance their usability in many fields not only for students.

Now that we have understood a lot about vedic maths, let us get going with vedic maths sutras and explore the immense potential of this method to make mathematics easy. After studying this method, the question "mathematics how to" would never arise!

Vedic maths sutras

Primary Sutras (formulas)

These vedic maths sutras deal with individual mathematical operations (i.e. multiplication, division, squares, factorization, etc.)

1. Last from 10 and the rest from 9
2. Cross-wise and Vertical
3. By one more than the one before.
4. Apply  after transposing
5. It is zero if the Samuccaya is Same
6. If One is in Ratio the Other is Zero
7. By Subtraction and by Addition
8. By the Completion or Non-Completion
9. Differential Calculus
10. Use Deficiency
11. Specific and General
12. The Ultimate and Twice the Penultimate
13. The Remainders by the Last Digit
14. By One Less than the One Before
15. All the Multipliers.
16. The Product of the Sum

Secondary Sutras (Formula)
These vedic maths sutras help augment the basic calculations while using the primary sutras or while doing normal mathematical calculations.

1. Proportionately
2. The Remainder Remains Constant
3. For 7 the Multiplicand is 143
4. The First by the First and the Last by the Last 
5. By Osculation
6. Lessen by the Deficiency
7. The Product of the Sum is the Sum of the Products
8. Set up the Square of the Deficiency
9. Whatever the Deficiency lessen by that amount and
10. Last Totalling 10
11. The Sum of the Products
12. Only the Last Terms
13. On the Flag (Not mentioned in vedic maths original books)
14. By Alternative Elimination and Retention
15. By Mere Observation

This is a complete list of vedic maths sutras what you can expect to learn from vedic mathematics and each and every vedic maths sutra has been discussed in detail in individual posts.

Mental Multiplication

Here we are talking about the sutra "Nikhilam" which would help us solve the mathematical multiplication in our mind without the use of calculator and even without having to write anything down! Let us understand the method with the help of examples.



Example 1:

605 x 504 = ?

We first need to select a base number (preferably a multiplier of 100 in case of three digit numbers to make it easier). In this case the base would be 500.

Now, define the distance of multipliers from the base as given below

605 -500 = +105 (the sign is very important)

504 -500 = +004

Now lets reproduce these numbers as:

605 + 105

504 + 004

From this, we derive two numbers and their names would be as follows:

LHS (is the result of multiplying the deficiencies) = +105 x +104 = +420

RHS (by adding diagonally on any one side) = 105 + 504 = 609 (remember, you can select any diagonal side, the answer will remain the same)

Now because our base is not a 10 series number (i.e. 10, 100, 1000, 10000, etc.) we need to establish the relation of the base with next highest 10 series number and the relation is 1000 / 2 = 500

Therefore, we have to express the LHS and RHS as follows:

(609 /2) | (+420)

304.5 + 420

304.5 | 420

Now see that the L.H.S has a decimal point ,so carry out that half to the R.H.S side from L.H.S.

Now R.H.S side becomes 420 + 500 = 920

304 | 920

Therefore ,605 x 504 = 304920

Example 2

485 x 475 = ?

The nearest base for the above two numbers are 500.So we fix this 500 as our base.Now, see that the above both numbers are less than the base.

How much deficient are are the above numbers are from base ?

485 -500 = -015

475 -500 = -025

485 - 015

475- 025

--------------
460 | ( +375) -->the RHS is got by adding diagonally on any one side ,the LHS is the result of multiplying the deficiencies .

230 | 375 ---- (Half of 460 is 230)

230 | 375

Therefore 485 x 475 = 230375

Example 3

614 x 495 =?.

The nearest base for the above two numbers are 500.So we fix this 500 as our base.Now, see that the above numbers .One is greater than the base and one is less than the base.

How much deficient are are the above numbers are from base ?

614 - 500 = +114

495 - 500 = -005

614 + 114

495 - 005

------------
609 | (-570)--->-->the RHS is got by adding diagonally on any one side ,the LHS is the result of multiplying the deficiencies .

(609 /2) | (-570)

304.5 | (-570) ----- > the minus sign is removed by 1000 -570 =430 and remove one from the L.H.S side

303.5 | 430

303| (430+500) ----> the 0.5 when carried to R.H.S, becomes 500 and add it to R.H.S

303| 930

Therefore ,614 x 495= 303930

These are general rules. There would not be issues when multiplications are above or higher, but when one is greater and one is lesser, follow the rules in the last example. All in all, the rules are:

The (0.5) on L.H.S is carried as 500 and added to the R.H.S every time you fall into a decimal (1/2).
The Minus signs on R.H.S is removed by adding 1000 to it.

It would seem very easy once you get in this mode of calculations.

Divisibility Testing

We all know divisibility tests for simple numbers like 2,3,4.etc but here I wish to demonstrate how to test for divisibility by 7, 11 and by higher prime numbers such as 13, 17, 19 and so on.

The Rule of 11
Let us first try out divisibility by 11 rule.

The first glance may force you to think that this method is complicated, but little time getting familiar with it, would make your life very simple.

For testing that a number is divisible by 11, add up two separate numbers. The first sum comprises the sum of the digits of the odd numbered columns of the number (1st, 3rd, 5th and so on). The second comprises the sum of the digits of the even numbered columns of the number (2nd, 4th, 6th and so on).

If the difference between these two numbers is zero, 11 or a multiple of 11, the number is divisible by 11.

Example:

Let us understand this with the example of number: 142,857.

Here, we now test 142,857 for divisibility by 11.

1 + 2 + 5 = 8

4 + 8 + 7 = 19

19 - 8 = 11

142,857 is divisible by 11.

The Rule of 7

So now we come to the divisibility test for the only number remaining, 7. When using this test, which can be applied to any other prime number, you only need to know the multiples of the candidate factor from zero and the factor itself through to five times the factor.

For example, in working out how to divide a number by 7, you only need to know 0, 7, 14, 21, 28 and 35. If you are testing for divisibility by 13, you only need to know 0, 13, 26, 39, 52 and 65.

Method:

The working behind this method is outlined below.

First of all, if a number a is a multiple of x:-

a = n1x

then any multiple b of a is also going to be a multiple of x:-

b = n2a ==> b = n2n1x

Next, any integer above 1 can be shown to be the sum of two smaller integers:-

c = d + e (d < c; e < c; d > 0; e > 0).

If we know that one of these two right hand terms is a known integer multiple of the candidate factor, we can eliminate it and test for divisibility with the smaller term, For example, substitute b for d:-

c = n2a + e

e = c - n2a

If, when we test e, we discover that it too is a multiple of x:-

e = n3x

we can deduce that the whole number c must be divisible by x.

This is the test: by eliminating known multiples of the candidate factor, if we obtain a final residue of 0 or the candidate factor, the number is divisible by the candidate factor.

To test this, we once again turn back to 142,857 and test it to see if it is divisible by 7. We already know that 142,857 is divisible by 11.

We can begin by eliminating the easiest multiples of 7. When eliminating a multiple, do not be afraid to take big bites.

142,857 - 140,000 = 2,857

2,857 - 2,800 = 57

Since the nearest multiple of 7 is now 56, eliminating that yields the final residue, 1. So 142,857 is not divisible by 7.

Let us test for divisibility by another prime factor, namely 17.

142,857 - 17 = 142,840

14,284 - 34 (2x17) = 14,250

1,425 + 85 (5x17) = 1,510

151 - 51 (3x17) = 100

100 - 85 = 15

Since the residue is neither 0 nor 17, 142,857 is not divisible by 17.

This technique can be applied to divisibility tests by any prime factor, and all it requires is the knowledge of the multiples of the candidate factor from 0 to 5x the factor.

Before wrapping up this article, I will conclude my business with 142,857. In my analysis of the candidate factors which go into 142,857 I determined that this number has the following factors, and only these factors:=

142,857 = 3 x 3 x 3 x 11 x 481

142,857 is an interesting number with connections to the number 7, which we shall discuss in a later post. However, 142,857 itself is not divisible by 7.

This rules are based on simple mathematical rules extrapolated to come up with a solution easier and practical for day to day issues. I hope you enjoyed knowing this!

Finding Squares

You know the squares of 30, 40, 50, 60 etc.

but if you are required to calculate square of 31 or say 61 then you will scribble on paper and try to answer the question.

Can it be done mentally?

Some of you will say may be and some of you will say may not be.

But if I give you a formula then all of you will say, yes! it can be.

What is that formula….. The formula is simple and the application is simpler.

Say you know 60sq = 3600

Then 61sq will be given by the following

61sq = 60sq + (60 + 61) = 3600 + 121 = 3721

or Say you know 25sq = 625 then 26sq = 625 + (25 + 26) = 676

Like above, you can find out square of a number that is one less than the number whose square is known.

Multiplication at your fingertips!

Here I am going to talk about "Ekadhikena Purveṇa" i.e "One more than the previous one" and very important Vedic Mathematics formula and how to use it for multiplication.

Application:

This formula can be applied to multiplication of numbers that satisfy both the following conditions

1. the same first digit
2. the sum of their last unit digits is 10.

We have already discussed about applying this formula for deriving squares for any number ending with 5, in previous post (Amazing way to derive squares within seconds). So here, I am going to talk about application of this formula for multiplication of numbers satisfying the conditions stated above.

Method:

Let us break the method into two parts:

1.  Deriving last two digits of answer
2. Deriving all previous digits of answer

1. Deriving last two digits of answer
These shall be multiplication of last digit of the numbers we are looking to multiply.

2. Deriving all previous digits of answer

All the digits in the multipliers other than the last would constitute base for us and let us call them B1.

Another base is B1 +1 (Because the formula is about one more than the previous one…)

So previous digits in the final answer would be simply [B1 x (B1+1)]

Example:

Let us understand this with an example: 24 x 26

Last two digit of the answer:

4 x 6 = 24 (If this answer is in single digit, then we need to put 0 before it)

For other digits, we need B1 and it goes as follows:

B1 = 2 (All the digits other than the last 5 in the number we want to derive square for)

B1 + 1 = 2 + 1 = 3

Therefore other digits of the answer preceding 25 are = 2 x 3 = 6

Hence the final answer would be 624.

Logic:

So what’s the logical explanation of this wonderful shortcut:

We all know that (a-b) x (a+b) = a²-b²

Now how does this help with multiplication of 24 & 26? It is like this:

24 x 26

(25-1) x (25+1)

25² – 1²

625 – 1 (25² derived from the same formula applied for squaring discussed in the previous post)

624 (Well this matches!)

Vedic mathematics rocks!

Amazing way to calculate squares within seconds

"Ekadhikena Purveṇa" as we discussed for division is the Sanskrit term for "One more than the previous one" and very important Vedic Mathematics formula. We already know from the previous post that this formula can be used for both multiplication and division, here let us have a look at how it helps for multiplication:
Application:
This formula can be applied to multiplication of numbers that satisfy both the following conditions

1. the same first digit
2. the sum of their last unit digits is 10.

Even though it might so appear that this formula is useful in limited cased but an interesting category of numbers that satisfy both above requirements are all the numbers ending with 5. So when we need to derive squares for any number ending with 5, this formula is applicable straightaway. In this post, I am going to talk about limited application of this formula for deriving squares of numbers with last digit as 5.

Method:

Let us break the method into two parts:

1. Deriving last two digits of answer
2. Deriving all previous digits of answer

1. Deriving last two digits of answer

No application of mind, last two digits are 25, always.

2. Deriving all previous digits of answer

All the digits in the number before the last five would constitute base for us and let us call them B1.

Another base is B1 +1 (Because the formula is about one more than the previous one…)

Digits previous of 25 in the final answer would be simply [B1 x (B1+1)]

Example:

Let us understand this with an example of squaring 25

Last two digit of the answer are 25

For other digits, we need B1 and it goes as follows:

B1 = 2 (All the digits other than the last 5 in the number we want to derive square for)

B1 + 1 = 2 + 1 = 3

Therefore other digits of the answer preceding 25 are = 2 x 3 = 6

Hence the final answer would be 625.

So now squaring is not a headache anymore having understood vedic mathematics, right?

Logic:

Learning just the shortcut method isn’t enough, so let us understand the logic how this operates. For explaining the logic we need to use the basic concepts of factorization:

We all know that (a+b)² = a² + 2ab + b² 

Now how does this square of numbers ending with 5 work in this case?

Let us continue with the above example:

25²

(20 + 5)²

20² + (2x20x5) + 5² (Factorization using the above rule)

20² + (20x10) + 5² (Simplified with multiplication of 5x2 for middle term)

20 (20 + 10) + 5² (taking 20 common from first two terms)

20 (30) + 5²

2 x 10 x 3 x 10 + 5²

(2x3) x (10x10) + 5²

(2x3) x (100) + 5²

Now this expression looks like this [B1 x (B1+1)] x 100 + 5²

Because in this expression two things are constant i.e. whatever is the B’s multiplication that shall at the end be multiplied by 100 and 25 shall be added so it is safe to make a rule that last two digits of the answer are going to be 25 and right preceding these digits we need to insert the multiplication of B1 & B1+1

Now you must feel awakened with the logical explanation of Vedic mathematics.

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Easier Solution for Complex Divisions

Now let us have a look at a sutra - "Ekadhiken Purven" out of a total of sixteen such sutras of Vedic Mathematics. Though this deals with multiplication and division both. Will talk about multiplication using this later, lets concentrate on division for the time being.

So first things first, when this sutra would apply

Values like 1/x9 (i.e. 1/19, 1/29, 1/39, etc.)

There are two methods by which we can approach this sutra I shall explain both in detail:

Prob: 1/19
First Conclusion:

19 is not a factor of either 2 or 5 which means that the result of this division is a purely circulating decimal.

Therefore, we must find out what are going to be the number of digits in this circulating decimal and the answer is 18 (divisor -1).

These 18 digits form two equal parts of 9 each and they complement each other to sum up to 9 as shown below: (1 denotes the left most digit of the result and 18 denotes the right most or last digit of the circulating sequence we shall receive out of this result)

1     2    3     4     5    6    7    8    9
10  11  12  13  14  15  16  17  18

This shows that the sum of digits at 1st place and 10th place in the circulating sequence should be 9 and so with sum of 2nd and 11th place and for all the rest of the pairs.

Thus, we effectively need to find out either 1st to 9th or 10th to 18th digits complement them respectively with 9 to get the other part of the answer.

There are two method in which the sutra can be used as shown below:

Method 1: Multiplication - To find out digits at places 10th to 18th
Method 2: Division - To find out digits at the places 1st to 9th

1. Multiplication (Deriving digits at places 10th to 18th of the answer)

This method gives answer from the right most digit (i.e. 18th) onwards to the left most digit (i.e. 10th)

18th Digit: Simply put the dividend (or numerator of the problem 1/19) here i.e. 1

17th to 10th Digits:

Just look at the problem as 1/x9 which will give us value of x=1, now add one to x which will make it 2 and thats our multiplier for the deriving the rest of the answers. so

Place  Digit
18th      1
17th      2 (i.e. 2 multiplier as explained above is multiplied with 1 i.e. 18th digit)
16th      4 (i.e. 2 multiplied with 17th digit)
15th      8 (i.e. 2 multiplied with 16th digit)
14th      6 (i.e. 2 multiplied with 15th digit, carry over 1)
13th      3 (i.e. 2 multiplied with 14th digit + carry over 1 = 13, so 1 carried over)
12th      7 (i.e. 2 multiplied with 13th digit)
11th      4 (i.e. 2 multiplied with 12th digit, carry over 1)
10th      9 (i.e. 2 multiplied with 16th digit + carry over 1 = 9)

So the set of digits from 10th to 18th place looks like this

9 4 7 3 6 8 4 2 1

Now lets subtract all of them from 9 to get 1st to 9th digits respectively;

9 4 7 3 6 8 4 2 1
0 5 2 6 3 1 5 7 8

So the final answer is:

1/19 = 0.052631578947368421 (circulating sequence)

2. Division (Deriving digits at places 1st to 9th of the answer)

This method gives answer from the left most digit (i.e. 1st) onwards to the right most digit (i.e. 9th)

Again, let us just look at the problem as 1/x9 which will give us value of x=1, now add one to x which will make it 2 and thats our magic divisor for the deriving the answer.

1st Place: a / b where a = dividend of the problem i.e. 1 and b = the magic divisor = 2

So, 1 / 2 is 0 with a remainder of 1. Hence, 1st place = 0

2nd place onwards: (in terms of a/b for each place)

Thus our 1st to 9th digits are

0 5 2 6 3 1 5 7 8

Now lets subtract all of them from 9 to get 1st to 9th digits respectively;

0 5 2 6 3 1 5 7 8

9 4 7 3 6 8 4 2 1

So the final answer is:

1/19 = 0.052631578947368421 (circulating sequence)

So with blessings of Vedic Mathematic, a little practice on this and you shall be quick like a gun!!

Will come back with application of this Vedic Maths sutra for multiplication which is even more interesting till then stay tuned and visit GyanCircle.com

Multiplication Made Easy Using The Line System

Now that we have seen the simpler way of multiplication using vedic mathematics for two digit numbers at Basics of Vedic Mathematics, I will show you an alternate method which even doesn't require multiplication of two single digit numbers to derive product of any large numbers.

Keep connected to www.gyancircle.com for more updates.

This is the fun of vedic mathematics where you can multiply without multiplying!

Basics of Vedic Mathematics

After introduction to Vedic Mathematics, let us start with basics thereof.
It would be good to take multiplication, and for that any 2 digit number with any other 2 digit number.

Basic Definitions:

Vedic mathematics is based on the concept of placing the numbers either at the unit place or tenths or hundredth and so on. So for ease of understanding, let us use this legend:

Unit Place: UP

Tenth Place: XP

Hundredth Place: HP

Thousand Place: TP

Ten thousandth Placce: TXP

Hundred thousandth Place: THP (and so on..)

Let us take an example of 23 x 45:

So now we need to place this numbers under each other in such a way that UP of both multipliers are in one column and XP numbers in one column:

XP UP

2 3

4 5

1. Now we have to start with UP and multiply both the numbers there. (i.e. 3 x 5) and the answer is 15. Out of this answer 5 would be placed on the UP of the product of two multipliers and 1 would be a carry over. So we now know that our final product has a UP of 5.

2. Second step is then to go for cross multiplication which means:

"XP of first multiplier" x "UP of Second multiplier" = 2 x 5 = 10

"XP of Second multiplier" x "UP of First multiplier" = 4 x 3 = 12

3. Now the sum of above two products added with carry over if any, would give us the XP of our final answer which is to be derived as follows:

Sum of Cross Multiplication (from Step 2) = 10 + 12 = 22

Carry over from the UP (from Step 1) = 1

XP for the final product = 23

As we did in step 1, number 3 shall occupy the XP of final product and 2 shall be a carry over. so the answer constructed by us so far look like this _ _ 3 5. Now we shall go into final steps to identify TP and TXP of the final product.

4. Multiplication of XP of both multipliers i.e. 2 x 4 = 8.

We need to add the carry over of two from step 3 into this which would give us 8 + 2 = 10. So 0 is the TXP and 1 is TP of the final product.

Thus, the final product looks like this 1035.

Benefits of this method

Now let us revisit and see what we have actually done to achieve this product of two digit numbers:

1. Multiplication of single digit number

2. Sum of two digit numbers.

Thus, the requirement of multiplying two digit numbers has become very very easy using this Vedic Maths technique and with a little bit of practice on this lines would make you very quick in deriving products.

Concept

Also, to understand the concept better, let us take this pictorial demonstration of steps. one this is understood, you shall be able to apply the concept to even larger digit numbers as well.


So Vedic Maths Rocks!

I hope this would help you master the trick for multiplication of two digit numbers and shortly I shall come up with multiplication of larger digit numbers using wonders of Vedic Mathematics.

As I mentioned previously, in this attempt to simplify the study and make it more interesting, I am supported by Gyancircle.com. Do visit them.